@Arjun The density of the air will certainly be decreased, though I don't know how much or if it makes a difference to the drone.

I would guess the updraft and turbulence above a fire would be the main problem.

I agree that the first statement is correct. But I do not understand how to interpret the second statement.

I believe the boundary surface for |psi|^2 gives the shape of the orbital?

I agree that it isn't very clear.

Is it specifically about atomic orbitals?

I believe so

I don't really get what the boundary surface for |psi| could imply

It's a surface where |ψ| is constant, but it is not clear how the question expects you to compare this with a surface where |ψ|² is constant.

Is it possible that the question models an s-orbital (where the boundary surface is a neat sphere)? From my understanding, the probability |ψ|^2 would be constant all over the surface ...

Is this condition valid for other orbital shapes? I don't think so.

|ψ|² is proportional to the electron density, so a surface of constant |ψ|² is also a surface of constant electron density. And this is what we normally take to mean the shape of the orbital.

So for all orbitals the shapes of the orbitals you see drawn in books are surfaces of constant |ψ|².

A surface of constant |ψ|² is also a surface of constant |ψ|, but it isn't clear what the question means by asking if the surfaces are different.

@JohnRennie I see

@JohnRennie yes i looked into it for a few more hours, the air is turbulent close to the fireplace, which causes imbalance and if the motors can't generate enough RPM to draw in sufficient airflow, it eventually crashes

it's a hard job to make a drone capable to fly into burning buildings

Huh, they mentioned nucleus

Alright so we won't consider copper electrons I guess

@JohnRennie Hi Sir! I was thinking, in formula for radial acceleration in polar coordinates, the term "r" in centripetal acceleration represents radius of curvature or radius vector (magnitude)?

𝑟 is the magnitude of the radius vector.

I had this doubt when i was studying kepler's laws, while deriving velocity at apogee/perigee, I equated the force of gravity on a planet to centripetal acceleration (in which i used , mv^2/ r, where r is position vector at apogee/perigee from one of the foci and v is the velocity of planet at apogee/perigee. I didn't get the answer, in fact the answer used R(radius of curvature at apogee/perigee) instead of radius vector.

But i believe, when we write centripetal acceleration in polar coordinates, r is radius vector...

@JohnRennie sir where am i getting wrong

It's because the centripetal acceleration at apogee and perigee is not equal to the gravitational acceleration.

At apogee the centripetal acceleration is less than the gravitational acceleration, and that's why after the apogee the body falls inwards i.e. 𝑟 decreases.

@JohnRennie but i believe only force acting is gravitational force? and velocity is perpendicular to radius vector?

At perigee the centripetal acceleration is greater than the gravitational acceleration, and that's why after the perigee the body moves outwards i.e. 𝑟 increases.

@JohnRennie sir, can you pls write the equations?

in polar form

What equations do you want me to write?

@JohnRennie for centripetal acceleration at apogee/perigee

And that's why in a circular orbit 𝑟 is constant.

But in an elliptical orbit 𝑟 is not constant and that means d²r/dt² must be non-zero.

Yes?

@JohnRennie umm..... sorry sir, but i didn't get it.... you say that radial acceleration = -GM/r² + v²/r, but shouldn't it be equal to gravitational acceleration???

i mean the net radial acceleration?

The net radial acceleration is just d²r/dt² i.e. telling us how the magnitude of the radial vector changes with time. Yes?

yes

And in a circular orbit the magnitude of the radial vector is constant, so both dr/dt and d²r/dt² are zero.

Yes?

yes

umm...

You are mixing up two different things.

sir i meant this...does d²r/dt² represent linear radial acceleration?

@JohnRennie sorry sir, help me to understand this

No, it just represents how the 𝑟 coordinate changes with time.

yes

But take a point to one side of the straight line and measure the 𝑟 coordinate from that point, then 𝑟 is not constant as the particle moves along the line. Yes?

@JohnRennie yes

So d²r/dt² is not the same as the acceleration 𝑎. It is just how the 𝑟 coordinate changes.

@JohnRennie i see

This is a surprisingly hard idea to get. I remember being puzzled by it myself.

@JohnRennie umm... so what should we use? wouldn't it depend upon the origin?

The equation you posted above gives the acceleration vector i.e. the rate of change of the velocity vector. It's the acceleration you would feel (the "g force") if you were the one moving.

In general that is not the same as d²r/dt²

@PinkAura It depends on what you are asking.

@JohnRennie umm... like?

@JohnRennie the image that i posted gives acceleration? but didn't they do d²r/dt²? to come to this

@JohnRennie sir can you link some sources to have my concepts cleared about this... it is really troubling me

I have to go I'm afraid. We'll need to pick this up tomorrow. Sorry :-(

@JohnRennie ok sir i'll ping you, thanks

@JohnRennie i think I get this, I misunderstood d²r/dt² , it is in fact how radius vector's "magnitude" changes(like you said), but can we determine this? if we know say, distance of apogee/perigee and masses, other essential things(like semi major and minor axes,etc.) but not the velocity at the apogee and perigee... Thanks sir